3 years ago

Complete each congruence statement by naming the corresponding angle or side.

Mathematics

1 answer:

alexandr1967 [171]3 years ago

5 0

The congruence statement should be: WV = XZ

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Find a cubic function with the given zeros. -6, 7, -4

Nookie1986 [14]

$f(x)=a(x-x_1)(x-x_2)(x-x_3)\\\\x_1=-6;\ x_2=7;\ x_3=-4\\\\f(x)=(x+6)(x-7)(x+4)=(x^2-x-42)(x+4)\\\\=x^3+3x^2-46x-168$

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3 years ago

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3 years ago

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You can make one square with four toothpicks. Show how you can make two squares with 10 toothpicks and five squares with 12 toot

dsp73

Answer:

by doing it the same way you did with the first one.

Step-by-step explanation:

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2 years ago

Cubed root x cubed root x2

Yuki888 [10]

Answer:

Final answer is $\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x$ .

Step-by-step explanation:

Given problem is $\sqrt[3]{x}\cdot\sqrt[3]{x^2}$ .

Now we need to simplify this problem.

$\sqrt[3]{x}\cdot\sqrt[3]{x^2}$

$\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}$

Apply formula

$\sqrt[n]{x^p}\cdot\sqrt[n]{x^q}=\sqrt[n]{x^{p+q}}$

so we get:

$\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{1+2}}$

$\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=\sqrt[3]{x^{3}}$

$\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x$

Hence final answer is $\sqrt[3]{x^1}\cdot\sqrt[3]{x^2}=x$ .

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3 years ago

5x+6y=7 need to find the slope intercept and standard form

Oksi-84 [34.3K]

The problem is already in standard form but to get it into slope intercept form you rewrite the equation so it looks like
6y=-5x+7 divide all by 6
y=-5/6x+7/6
slope being -5/6 and b being 7/6

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3 years ago